中學數學

高中數學

數列級數

\(1 + 2 + \dots + n = \frac{n(n + 1)}{2}\)
\(1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6} = \frac{n(n + \frac{1}{2})(n + 1)}{3}\)
\(1^3 + 2^3 + \dots + n^3 = \frac{n(n + 1)}{2}^2\)

指對數

任意數的0次方為1
0的非零任意數次方為0
0的零次方通常為undefine（黑洞？）

設\(a > 0, b > 0, m , n \in \Bbb {R}\)

指數率

\(a^m \cdot a^n = a^{m + n}\)
\(a^{m^n} = a^{mn}\)
\((ab)^n = a^n b^n\)
\(a^{-n} = \frac{1}{a^n}\)
\(a^{\frac{m}{n}} = \sqrt[n]{a^m}\)

\(\log_a 1 = 0\)
\(\log_a a = 1\)
\(\log_a mn = \log_a m + \log_a n\)
\(\displaystyle \log_a \frac{m}{n} = \log_a m - \log_a n\)
\(\log_a x^m = m \log_a x\)
\(\log_a x = \frac{1}{\log_x a}\)
換底公式 \(\log_a{b}=\frac{{{\log }_{c}}b}{{{\log }_{c}}a} (c>0,\ c\ne 1) \)

三角函數

\(\sin(-x) = -\sin{x}\)
\(\cos(-x) = \cos{x}\)
\(\sin(\frac{\pi}{2} - x) = \cos{x}\)
\(\cos(\frac{\pi}{2} - x) = \sin{x}\)
\(\sin(\frac{\pi}{2} + x) = \cos{x}\)
\(\cos(\frac{\pi}{2} + x) = -\sin{x}\)
\(\sin(\pi - x) = \sin{x}\)
\(\cos(\pi - x) = - \cos{x}\)
\(\sin(\pi + x) = - \sin{x}\)
\(\cos(\pi - x) = - \cos{x}\)

\(\sin^2 {x} + \cos^2 {x} = 1\)
\(\tan^2 {x} + 1 = \sec^2 {x}\)
\(\cot^2 {x} + 1 = \csc^2 {x}\)

和差化積

\(\sin(\alpha + \beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha\)
\(\sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha\)
\(\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)
\(\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \)
\(\displaystyle \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}\)
\(\displaystyle \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}\)

倍角半倍角

\(\sin{2 \alpha} = 2 \sin \alpha \cos \alpha\)
\(\cos{2 \alpha} = \cos^2 {x} - \sin^2 {x} = 2 \cos^2 {x} - 1 = 1 - 2 \sin^2 {x}\)
\(\displaystyle \tan{2 \alpha} = \frac{2 \tan \alpha}{1 - \tan^2 \alpha}\)
\(\displaystyle \sin{\frac{\alpha}{2}} = \pm \sqrt{\frac{1 - \cos \alpha}{2}}\) (determine whether it is + or - by finding the quadrant that \(\frac{\alpha}{2}\) lies in)
\(\displaystyle \cos{\frac{\alpha}{2}} = \pm \sqrt{\frac{1 + \cos \alpha}{2}}\) (same as above)
\(\displaystyle \tan{\frac{\alpha}{2}} = \frac{1 - \cos \alpha}{\sin \alpha} = \frac{\sin \alpha}{1 + \cos \alpha}\)